Page 154 - Advanced engineering mathematics

134 CHAPTER 4 Series Solutions

and so on. In general,
1
c n = c 0
n!(n + 1)!
for n = 1,2,3,···, and one Frobenius solution is
∞
1
n+1
x
y 1 (x) = c 0
n!(n + 1)!
n=0
1 1 1

2
3
4
= c 0 x + x + x + x + ··· .
x 12 144
For a second solution, put r = 0 into the general recurrence relation to obtain
n(n − 1)c n − c n−1 = 0
for n =1,2,···. If we put n =1 into this, we obtain c 0 =0, violating one of the conditions for the
method of Frobenius. Here we cannot obtain a second solution as a Frobenius series. Theorem
4.2, Case (4), tells us to look for a second solution of the form
∞

n
∗
y 2 (x) = ky 1 ln(x) + c x .
n
n=0
Substitute this into the differential equation to obtain

∞
1 1 n−2

x ky ln(x) + 2ky 1 − ky 1 + n(n − 1)c x
∗
n
1
x x 2
n=2
∞

n
∗
− ky 1 ln(x) − c x = 0.
n
n=0
Now

k ln(x)[xy − y 1 ]= 0,
1
because y 1 is a solution. For the remaining terms, let c 0 = 1in y 1 (x) for convenience (we need
only one more solution) to obtain
∞ ∞ ∞ ∞
1 1
n
n
n
∗
2k x − k x + c n(n − 1)x n−1 − c x = 0.
∗
n
n
(n!) 2 n!(n + 1)!
n=0 n=0 n=2 n=0
Shift indices in the third summation to write
∞
1 n 1 n
2k x − k ∞ x
(n!) 2 n!(n + 1)!
n=0 n=0
∞ ∞

n n
∗
+ c n+1 n(n + 1)x − c x = 0.
n
n=1 n=0
Then
∞
2k k
∗ 0 ∗ ∗ n
(2k − k − c )x + − + n(n + 1)c − c x = 0.
0 2 n+1 n
(n!) n!(n + 1)!
n=1
This implies that k − c = 0, so
∗
0
∗
k = c .
0
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October 14, 2010 14:17 THM/NEIL Page-134 27410_04_ch04_p121-136